Two lightweight rods 24 cm in length are mounted perpendicular to an axle and at 180 ∘ to each other. At the end of each rod is a 480-g mass. The rods are spaced 42 cm apart along the axle. The axle rotates at 5.0 rad/s. (a) What is the component of the total angular momentum along the axle? (b) What angle does the vector angular momentum make with the axle? [Hint: Remember that the vector angular momentum must be calculated about the same point for both masses, which could be the cm.]

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Answer:

Explanation:

Angular momentum ( L ) = moment of inertia x angular velocity ( I X ω )

Moment of inertia of two  480 g masses about axle = 2 x mr²  = 2 x 480 x10⁻³ x( 24 x 10 ⁻ 2 )²  = 0. 552960  kg m².

Angular velocity = 5 rad / s.

Angular momentum = 0.552960 x 5 = 2.765 kg m2.

The direction of angular momentum will be along axle.So vector angular

momentum  makes zero degree  with axle.

The component of the total angular momentum and the angle the vector angular momentum makes with the axle is mathematically given as

a) L= 2.765 kg m^2

b) [tex]\theta =0 \textdegree[/tex]

What are the component of the total angular momentum along the axle and the angle the vector angular momentum makes with the axle?

Question Parameters:

Generally, the equation for the Angular momentum   is mathematically given as

Angular momentum = moment of inertia x angular velocity

Therefore

L=i*w

Where Moment of inertia is

i= 2 x mr^2

i= 2 x 480 x10^{-3} x( 24 x 10 ⁻ 2 )^2  

i= 0. 552960  kg m^2.

Therefore

L= i*w

L  = 0.552960 x 5

L= 2.765 kg m^2

In conclusion, Due to the fact that the angular momentum won't change direction but maintain its movement along the axle the angle the vector angular momentum makes will be zero.

[tex]\theta =0[/tex]

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